If $x \dagger y = 3x-8y$ and $x \star y = 4x^{2}+y^{2}$, find $-2 \dagger (0 \star -1)$.
Explanation: First, find $0 \star -1$ $ 0 \star -1 = 4(0^{2})+(-1)^{2}$ $ \hphantom{0 \star -1} = 1$ Now, find $-2 \dagger 1$ $ -2 \dagger 1 = (3)(-2)-(8)(1)$ $ \hphantom{-2 \dagger 1} = -14$.